# Thread: Conjugacy

1. ## Conjugacy

Hi everyone,
Let G be a group acting on itself by conjugation,

Describe the set-theorem and numerical forms of the Class Equation for these actions explicitly when
(i)
(ii)

What do they mean by set theorem and numerical forms? Is that reflective, transitive - all the relations?

I think to do this question we have to compute each of element in the group, it is easy to compute the identity element.In particular, can you show me how to compute one element in the group?

Thanks for your time

2. Originally Posted by knguyen2005
Hi everyone,
Let G be a group acting on itself by conjugation,

Describe the set-theorem and numerical forms of the Class Equation for these actions explicitly when
(i)
(ii)

What do they mean by set theorem and numerical forms? Is that reflective, transitive - all the relations?

I think to do this question we have to compute each of element in the group, it is easy to compute the identity element.In particular, can you show me how to compute one element in the group?

Thanks for your time
I am not sure what the problem is asking, I can show you have to get the conjugacy class equations if that would help i.e. finding the orbits of this group action.

3. Yeah, I meant find the orbits of the group. How do you do it?
Thanks

4. Originally Posted by knguyen2005
Yeah, I meant find the orbits of the group. How do you do it?
Thanks
I will do $A_4$ for you. First of all if two elements are conjugate in $A_4$ then they are definitely conjugate in $S_4$, therefore two permutations in $A_4$ are conjugate then they must be the same type of permutation. The permutations of $A_4$ are: id, (12)(34),(13)(24),(14)(23),(123),(132),(124),(142) ,(134),(143),(234),(243).

The orbit of each of these elements is the conjugacy classes of that element. Of course, $[\text{id}] = \{ \text{id}\}$ since the identity is only conjugate to itself. Let us pick another element, say $(12)(34)$. Now the number of elements in $[(12)(34)]$, the conjugacy class of (12)(34), is equal to the index of the centralizer of that element in the group ( $A_4$) - this result follows by the orbit-stabilizer theorem. Notice that $C(12)(34)$, the centralizer, contains $\{ \text{id},(12)(34),(13)(24),(14)(23)\}$, thus there are at least 4 elements in the centralizer. By Lagrange's theorem $C(12)(34) = 4 \text{ or }6$, the reason why it cannot be $12$ is because the center of $S_4$ (and certainly $A_4$) is trivial. It cannot be $6$ because otherwise $[G:C(12)(34)]=2$ which would mean the conjugacy class of $(12)(34)$ would consist of another element of the same shape, so either $[(12)(34)]=\{(12)(34),(13)(24)\}$ or $[(12)(34)]=\{(12)(34),(14)(23)\}$. In either case the remaining 2-2 cycle must be only conjugate to itself i.e. it lies in the center, which is impossible because the center is trivial. Thus, $[G:C(12)(34)] = 3$ and so $[(12)(34)] = \{ (12)(34),(13)(24),(14)(23)\}$. Of course, we used over here an argument to compute $C(12)(34)$, we did not need to, you can do it directly by working out the computations but I chose not to do it that way because it is a little boring.

We are left with only the 3-cycles. Take $(123)$, notice that $\left< (123) \right> \subseteq C(123)$ therefore $3$ divides the order of $C(123)$ but order of $C(123)$ divides $12$ therefore $|C(123)| = 3\text{ or }6$. Let us compute the centralizer. Of course we do not actually want to do any computation because we are mathematicians and we are lazy, so let us think of an argument that will tell us what the order of the centralizer must be. Notice that if a 3-cycle outside $\left<(123)\right>$ is contained then its inverse must be contained also, therefore $C(123)$ would have at least 5 elements, but since it cannot have 5 elements it would have to contain another elements. If this other element was a 2-2-cycle then the product of this 3-cycle and 2-2-cycle would produce another distinct element and so the centralizer would have at least 7 elements, a contradiction; if other elements would be a 3-cycle then its inverse must also be contained with means the centralizer has at least 7 elements, a contradition. The other case is that a 2-2-cycle is contained in the centralizer, but if a 2-2-cycle is contained then it means the cycle has at least 4 elements, which means it has to contain another 2-2-cycle (it cannot be a 3-cycle by above) but that still leave 5 elements, so we have shown that if $C(123)$ is not $\left< (123)\right>$ then it must contain all the 2-2-cycles. A simple computation (I failed to avoid all computations) with any 2-2-cycle show that a 2-2-cycle does not compute with $(123)$ and so it forces $|C(123)| = 3$. Now $[G:C(123)] = 3$ which means $[(123)]$ consists of 3 elements.

Can you finish this problem?